linear algebra 08/14/2017 lecture 7 & 8

lecture 7:
the rank of the matrix. 矩阵的秩

do elimination, find the pivot columns and free columns.
pivot variables + free variables = n (column number)

reduced row echelon form 是一种矩阵的继续简化

求解null space
解法1：produce null space matrix R*N=0
R是已经被简化的矩阵，变成[I,F;0;0] I 是单元矩阵
N=[-F;I]

解法2：求解null space.把free variable 其中一个设为1，其余free variable 为0. 然后求出一组解，

lecture 8:
Ax = b
solvable when b is in C(A)
If a combination of rows of A gives 0 rows,

to find a compete solutions

1.x particular (找一个特殊解)： set all free variables as 0
2. 再把其中一个free variable 换成1，其余的为0，求几组特殊解


The case of full column rank 
r = n

what that imply about the solutions? What is that mean?
No free variables
=> N(A) = {0} null space only have 零向量
solution to Ax = b
x = unique solution, if exists (0 or 1 solution)

full row rank means 

r = m
can solve Ax = b for every b.
free variables n -r = n-m 个
肯定可以得出一个最简矩阵R=[I F]

special case: r = m = n

R = I
=> N(A) = {0} null space only have 零向量

Ax = b

r= m < n 

R= [I;0] => 0  or 1 solution

r = m<n

R = [I F] => infinite solution

r<m, r<n
R = [I,F;0,0]
=> 0 or infinite solutions